Problem: Find the limit as $x$ approaches negative infinity. $\lim_{x\to-\infty}\dfrac{5x^2+6x}{\sqrt{16x^4-5x^2}}=$
Explanation: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the numerator is $x^2$, let's divide by $x^2$. In the denominator, let's divide by $\sqrt{x^4}$, since for any value, $x^2=\sqrt{x^4}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{5x^2+6x}{\sqrt{16x^4-5x^2}} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{5x^2+6x}{x^2}}{\dfrac{\sqrt{16x^4-5x^2}}{\sqrt{x^4}}} \gray{\text{Divide sides by }x^2=\sqrt{x^4}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}\dfrac{\dfrac{5\cancel{x^2}}{\cancel{x^2}}+\dfrac{6\cancel x}{\cancel x \cdot x}}{\sqrt{\dfrac{16\cancel{x^4}}{\cancel{x^4}}-\dfrac{5\cancel {x^2}}{\cancel {x^2}\cdot x^2}}} \\\\ &=\lim_{x\to-\infty}\dfrac{5+\dfrac{6}{x}}{\sqrt{16-\dfrac{5}{x^2}}} \\\\ &=\lim_{x\to-\infty}\dfrac{5+0}{\sqrt{16-0}} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{5}{\sqrt{16}} \\\\ &=\dfrac{5}{4} \end{aligned}$ In conclusion, $\lim_{x\to-\infty}\dfrac{5x^2+6x}{\sqrt{16x^4-5x^2}}=\dfrac{5}{4}$.